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STD 11- 2. Relation and Function — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11- 2. Relation and Function4 Marks
MCQ
Domain of the function $f(x) = \sqrt {2 - 2x - {x^2}} $ is
  • A
    $ - \sqrt 3 \le x \le \sqrt 3 $
  • ✓
    $ - 1 - \sqrt 3 \le x \le - 1 + \sqrt 3 $
  • C
    $ - 2 \le x \le 2$
  • D
    $ - 2 + \sqrt 3 \le x \le - 2 - \sqrt 3 $
✓

Answer

Correct option: B.
$ - 1 - \sqrt 3 \le x \le - 1 + \sqrt 3 $
b
(b) The quantity under root is positive, when

If we want the range of $f(x)$ to be real then, $0 \leq 2-2 x-x^{2}$

$\Longrightarrow x^{2}+2 x \leq 2$

$\Longrightarrow x^{2}+2 x+1 \leq 3$

$\Longrightarrow(x+1)^{2} \leq 3$

$\Longrightarrow-\sqrt{3} \leq x+1 \leq \sqrt{3}$

$\Longrightarrow-\sqrt{3}-1 \leq x \leq \sqrt{3}-1$

So the domain of $x$ is $[-1-\sqrt{3},-1+\sqrt{3}]$

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